A tangent galvanometer has a coil of 25 turns | vedclass

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Question

(A) The formula for the current $i$ in a tangent galvanometer is given by $i = \frac{2r B_H}{\mu_0 N} 	an 	heta$.Given values are:Number of turns $N

Vedclass · Accepted Answer

$0.29$

Vedclass · Answer

(A) The formula for the current $i$ in a tangent galvanometer is given by $i = \frac{2r B_H}{\mu_0 N} 	an 	heta$.Given values are:Number of turns $N = 25$Radius $r = 15 \, cm = 0.15 \, m$Horizontal component of earth's magnetic field $B_H = 3 	imes 10^{-5} \, T$Deflection $	heta = 45^o$Permeability of free space $\mu_0 = 4\pi 	imes 10^{-7} \, T \cdot m/A$Substituting these values into the formula:$i = \frac{2 	imes 0.15 	imes 3 	imes 10^{-5}}{4\pi 	imes 10^{-7} 	imes 25} 	imes 	an 45^o$$i = \frac{0.9 	imes 10^{-5}}{100\pi 	imes 10^{-7}} 	imes 1$$i = \frac{0.9 	imes 10^{-5}}{\pi 	imes 10^{-5}} = \frac{0.9}{3.14159} \approx 0.2865 \, A$Rounding to two decimal places,we get $i \approx 0.29 \, A$.

$A$ tangent galvanometer has a coil of $25$ turns and a radius of $15 \, cm$. The horizontal component of the earth's magnetic field is $3 \times 10^{-5} \, T$. The current required to produce a deflection of $45^o$ in it is.....$A$

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